Question:medium

A metallic wire of density $\rho$ is placed horizontally on the surface of a liquid with surface tension $T$. What is the maximum radius the wire can have so that it remains supported by surface tension?

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Remember that surface tension acts on both sides of the wire length, effectively doubling the supporting force to $2TL$.
Updated On: Jun 6, 2026
  • $\sqrt{\frac{2T}{\pi \rho g}}$
  • $\sqrt{\frac{T}{\pi \rho g}}$
  • $\sqrt{\frac{3 \rho g}{2T}}$
  • $\sqrt{\frac{2 \pi}{T \rho g}}$
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The Correct Option is A

Solution and Explanation

Step 1: The balance that holds the wire up.
The wire floats on the liquid because the upward pull of surface tension balances the downward pull of gravity (its weight). At the largest possible radius these two are exactly equal.

Step 2: Write the weight.
Take a wire of length $L$ and radius $r$, made of material with density $\rho$. Its volume is $\pi r^2 L$, so its weight is \[ W = (\pi r^2 L)\rho g \]

Step 3: Write the surface tension force.
Surface tension acts along the line where the liquid surface meets the wire, on both sides of the wire. The total length of contact is $2L$, so the upward force is \[ F = 2 T L \]

Step 4: Set them equal.
\[ 2 T L = (\pi r^2 L)\rho g \] The length $L$ cancels from both sides: \[ 2 T = \pi r^2 \rho g \]

Step 5: Solve for the radius.
\[ r^2 = \frac{2T}{\pi \rho g} \;\Rightarrow\; r = \sqrt{\frac{2T}{\pi \rho g}} \]

Step 6: Conclusion.
The maximum radius the wire can have and still be held up is \[ \boxed{\,r = \sqrt{\dfrac{2T}{\pi \rho g}}\,} \]
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